Answer
The identity is verified.
$\frac{sin~x±sin~y}{cos~x+cos~y}=tan\frac{x±y}{2}$
Work Step by Step
$\frac{sin~x+sin~y}{cos~x+cos~y}=\frac{2~sin(\frac{x+y}{2})~cos(\frac{x-y}{2})}{2~cos(\frac{x+y}{2})~cos(\frac{x-y}{2})}=\frac{sin(\frac{x+y}{2})}{cos(\frac{x+y}{2})}=tan\frac{x+y}{2}$
$\frac{sin~x-sin~y}{cos~x+cos~y}=\frac{2~cos(\frac{x+y}{2})~sin(\frac{x-y}{2})}{2~cos(\frac{x+y}{2})~cos(\frac{x-y}{2})}=\frac{sin(\frac{x-y}{2})}{cos(\frac{x-y}{2})}=tan\frac{x-y}{2}$
